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Chapter 8: Problem 41

A certain adjustment to a machine will change the length of the parts it makes but will not affect the standard deviation. The length of the parts is normally distributed, and the standard deviation is 0.5 \(\mathrm{mm}\). After an adjustment is made, a random sample is taken to determine the mean length of the parts now being produced. The resulting lengths are as follows: $$\begin{array}{llllllllll} \hline 75.3 & 70.0 & 75.0 & 77.0 & 75.4 & 76.3 & 77.0 & 74.9 & 76.5 & 75.8 \\ \hline \end{array}$$ a. What is the parameter of interest? b. Find the point estimate for the mean length of all parts now being produced. c. Find the 0.99 confidence interval for \(\mu\)

Short Answer

Expert verified
a. The parameter of interest is the mean length of the parts (\(\mu\)). b. The point estimate of the mean length is calculated to be 75.32 mm. c. The 0.99 CI for \(\mu\) can be found using the formula \(\mu \pm Z_{\frac{\alpha}{2}}*\frac{\sigma}{\sqrt{n}}\) which leads to (74.36 mm, 76.28mm).

Step by step solution

01

Identify the Parameter of Interest

The parameter of interest will be the mean length of the parts after the machine adjustment, which is represented as \(\mu\).
02

Calculate the Point Estimate

To get the point estimate, you need to take the average of the observed data. This can be computed as: \(\mu = \frac{75.3 + 70.0 + 75.0 + 77.0 + 75.4 + 76.3 + 77.0 + 74.9 + 76.5 + 75.8}{10}\).
03

Calculate the 0.99 Confidence Interval for \(\mu\)

The formula for the Confidence Interval (CI) for the mean can be calculated as: \(\mu \pm Z_{\frac{\alpha}{2}}*\frac{\sigma}{\sqrt{n}}\). Where \(Z_{\frac{\alpha}{2}} = 2.576\) for a 0.99 CI, \(\sigma = 0.5mm\) (standard deviation given), and \(n = 10\) (number of observations). The margin of error equals to \(2.576 * \frac{0.5}{\sqrt{10}}\) and the confidence interval will be \(\mu\) plus minus the margin of error.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate
A point estimate is a single value that serves as a best guess or approximation for an unknown population parameter. In many statistical problems, like the one described in the exercise, we are interested in estimating the mean length of parts produced after a machine adjustment. To find this point estimate, you would calculate the mean of the sample you have. For instance, if you have the series of measurements like 75.3 mm, 70.0 mm, and so on, you simply add all these numbers up and divide by the total number of measurements. This calculation provides a single value that serves as the estimate of the population mean. It's important because it gives us a starting point for other statistical evaluations, such as confidence intervals.
Parameter of Interest
The parameter of interest is a critical concept in statistics. It refers to a specific quantity that we want to learn about in a population based on our sample data. In the context of this exercise, the parameter of interest is the true mean length of the produced parts represented by the symbol \( \mu \). It's the characteristic that the analyst wants to draw conclusions about. Knowing the parameter of interest helps in focusing analyses on what matters. The parameter of interest usually guides how data is collected and which statistical methods are appropriate to use. In this example, understanding that the mean length is the parameter of interest informs us that estimating the mean from our sample data accurately is vital for making further decisions.
Normal Distribution
Normal distribution, sometimes called the Gaussian distribution, is a probability distribution that is symmetric around the mean. In this context, the lengths of the parts produced by the machine are said to be normally distributed. The significance of normal distribution lies in its shape, which is characterized by a bell curve. Most of the observed data tends to cluster around the central mean, creating a symmetrical distribution.Understanding that the data follows a normal distribution allows us to make certain predictions and apply statistical methods such as confidence intervals. For example, it enables using established statistical values such as the Z-score \( Z_{\frac{\alpha}{2}} \) when calculating confidence intervals. The properties of normal distribution help in making inferences about the population from the sample data efficiently and accurately.

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Most popular questions from this chapter

The manager at Air Express believes that the weights of packages shipped recently are less than those in the past. Records show that in the past, packages have had a mean weight of 36.5 lb and a standard deviation of 14.2 lb. A random sample of last month's shipping records yielded the following 64 data values: $$\begin{array}{llllllll} \hline 32.1 & 41.5 & 16.1 & 8.9 & 36.2 & 12.3 & 28.4 & 40.4 \\ 45.5 & 15.2 & 26.5 & 13.3 & 23.5 & 33.7 & 18.3 & 16.3 \\ 15.4 & 39.7 & 50.3 & 14.8 & 44.4 & 47.7 & 45.8 & 52.3 \\ 48.4 & 10.4 & 59.9 & 5.5 & 6.7 & 17.1 & 20.0 & 28.1 \\ 48.1 & 29.5 & 22.9 & 47.8 & 24.8 & 20.1 & 40.1 & 12.6 \\ 24.3 & 43.3 & 32.4 & 57.7 & 42.9 & 36.7 & 15.5 & 46.4 \\ 51.3 & 38.6 & 39.4 & 27.1 & 55.7 & 37.7 & 39.4 & 55.5 \\ 26.9 & 15.7 & 32.3 & 47.8 & 33.2 & 29.1 & 31.1 & 34.5 \\ \hline \end{array}$$ Is this sufficient evidence to reject the null hypothesis in favor of the manager's claim? Use \(\alpha=0.01\)

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The average volunteer ambulance member is 45 years old and has 8 years of service, according to the Democrat \& Chronicle article "Unpaid ambulance workers could get 'pension"' (January 23,2005 ). The quoted statistics were based on the Penfield Volunteer Ambulance Squad of 80 members. If the Penfield Volunteer Ambulance Squad is considered representative of all upstate New York volunteer ambulance squads, determine a \(95 \%\) confidence interval for the mean age of all volunteer ambulance members in upstate New York. Assume the population standard deviation is 7.8 years.

Suppose you want to test the hypothesis that "the mean salt content of frozen 'lite' dinners is more than \(350 \mathrm{mg}\) per serving." An average of \(350 \mathrm{mg}\) is an acceptable amount of salt per serving; therefore, you use it as the standard. The null hypothesis is "The average content is not more than \(350 \mathrm{mg} "(\mu=350) .\) The alternative hypothesis is "The average content is more than \(350 \mathrm{mg} "\) \((\mu > 350)\) a. Describe the conditions that would exist if your decision results in a type I error. b. Describe the conditions that would exist if your decision results in a type II error.
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